New Year and Old Property CodeForces

本文介绍了一个算法问题的解决方案，该问题是计算给定区间内所有具有恰好一个二进制零的整数数量。通过预先计算特定的数字序列，并使用循环遍历的方式找出符合条件的数。

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参考博客：http://blog.youkuaiyun.com/loy_184548/article/details/50864501
PS：网上还有说用dfs跑寻找唯一具有一个0的情况，（貌似是官方答案，是模拟+dfs的形式，….（有意向的同学可以自己查一下））

The year 2015 is almost over.

Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510 = 111110111112. Note that he doesn't care about the number of zeros in the decimal representation.

Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?

Assume that all positive integers are always written without leading zeros.

Input

The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 1018) — the first year and the last year in Limak's interval respectively.

Output

Print one integer – the number of years Limak will count in his chosen interval.

Example
Input

5 10

Output

2

Input

2015 2015

Output

1

Input

100 105

Output

0

Input

72057594000000000 72057595000000000

Output

26

Note

In the first sample Limak's interval contains numbers 510 = 1012, 610 = 1102, 710 = 1112, 810 = 10002, 910 = 10012 and 1010 = 10102. Two of them (1012 and 1102) have the described property.

题目内容：给你一个区间[a,b]，求在[a,b]范围内的数的二进制编码只有一个零的情况。
我没有思路，参考博客的思路是求出二进制都为1的情况下的数字，在跑循环减掉一个0 ，-2^n方，每一次求出来的数必定符合题意，即可

#include<bits/stdc++.h>
using namespace std;
typedef long long  ll;
ll top[100];
inline void  find_top()
{
    int i;
    top[0]=1;
    for (i=1;i<=64;i++)
    {
        top[i]=top[i-1]+pow(2,i);
    }
}
int solve (ll &a,ll &b)
{
    int i,j;
    int ans =0;
    for (i=0;i<64;i++)
    {
        ll temp=pow(2,i);
        for (j=i+1;j<64;j++)
        if (top[j]-temp>=a&&top[j]-temp<=b)
            ans++;
    }
    return ans;
}

int main ()
{
    find_top();
    ll a, b;
    cin>>a>>b;
    cout<<solve(a,b)<<endl;


    return 0;
}